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A charged particle enters in a uniform m...

A charged particle enters in a uniform magnetic field perpendicular to it. Now match the following two columns.

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The correct Answer is:
`A to p; B to q; C to q; D to q,s`
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A charged particle enters a uniform magnetic field with velocity v_(0) = 4 m//s perpendicular to it, the length of magnetic field is x = ((sqrt(3))/(2)) R , where R is the radius of the circular path of the particle in the field. Find the magnitude of charge in velocity (in m/s) of the particle when it comes out of the field.

Motion of a charged particle in a uniform magnetic field |Questions

Knowledge Check

  • A charged particle enters a uniform magnetic field perpendicular to it. The magnetic field

    A
    increases the `K.E.` of the particle
    B
    decreases the `K.E.` of the particle
    C
    changes the direction of motion of the particle
    D
    does both (a) and ( c)

  • A charged particle enters into a uniform magnetic field with velocity v_(0) perpendicular to it , the length of magnetic field is x=sqrt(3)/(2)R , where R is the radius of the circular path of the particle in the field .The magnitude of change in velocity of the particle when it comes out of the field is

    A
    `2v_(0)`
    B
    `v_(0)//(2)`
    C
    `(sqrt(3)v_(0))/(2)`
    D
    `v_(0)`

  • When a charged particle enters a uniform magnetic field its kinetic energy

    A
    remains constant
    B
    increased
    C
    decreases
    D
    becomes

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